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In the given figure, ∆ODC~∆OBA, ∠BOC = 115° and ∠CDO = 70° . Find(i) ∠DCO (ii) ∠DCO (iii) ∠OAB (iv) ∠OBA. |
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Answer» (i) It is given that DB is a straight line. Therefore, ∠DOC + ∠COB = 180° ∠DOC = 180° − 115° = 65° (ii) In ∆ DOC, we have: ∠ODC + ∠DCO + ∠DOC = 180° Therefore, 70° + ∠DCO + 65° = 180° ⇒ ∠DCO = 180 − 70 − 65 = 45° (iii) It is given that ∆ ODC - ∆ OBA Therefore, ∠OAB = ∠OCD = 45° (iv) Again, ∆ ODC- ∆ OBA Therefore, ∠OBA = ∠ODC = 70° |
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